4. Let each side of the square that is cut out be s m
then the box will have dimensions s by 2-2s by 3 - 2s
and the
volume = s(2-2s)(3-2s)
expand, find the derivative, set that equal to zero and solve that quadratic
This question is probably used more than any other to introduce the concept of maximum/minimum by most textbooks, only the numbers will differ.
5. another straight-forward Calculus question
let the radius be r, and the height h cm
volume = πr^2h
πr^2h = 64
h = 64/(πr^2)
a) SA = one circle + the "sleeve" of the cylinder
= πr^2 + 2πrh
= πr^2 + 2πr(64/(πr^2))
= πr^2 + 128/r
d(SA)/dr = 2πr - 128/r^2 = 0 for a max/min of SA
2πr = 128/r^2
r^3 = 64/π
take cube root, sub back into h = ....
I actually have two questions:
4. An open box is to be made from a rectangular piece of material 3m by 2m by cutting a congruent square from each corner and folding up the sides. What are the dimensions of the box of the largest volume made this way, and what is the volume?
5. A cylindrical container w/ a circular base is to hold 64 cubic cm. Find its dimensions so that the amt (surface area) of metal required is a minimum when the container is
a. an open cup and
b. a closed can.
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W/ 4, I have no idea how to approach. All I got is that volume of the box would be s(s-2)(s-3), I think.
As for 5, I know the formula for the volume and surface area, but my question is about the open cup for a. You probably have to subtract something from somewhere...but where? I don't really know.
Thank you very much!
1 answer